ES>Work in n-dimensional space with basis e1, ... , en. You are given n-1 vectors, the rows of your matrix, by hypothesis linearly independent. The 1-dimensional null space is generated by a vector orthogonal to these. Use the n-dimensional generalization of the cross product. Append to your matrix the row [e1, ... , en]. Now you have an nxn matrix. Calculate its determinant, and express it as A1 e1 + A2 e2 + ... + An en. Then [A1, ... , An] is a basis for the null space. -- Gene ________________________________ From: Scott Fenton <sctfen@gmail.com> To: math-fun <math-fun@mailman.xmission.com> Sent: Saturday, July 13, 2013 4:48 PM Subject: [math-fun] Symbolic null space solutions Hi all, I'm not sure if Google's and my linear algebra education are failing me, but I've got a matrix problem that I have no idea how to solve. The basics go like this. I've got an n by (n-1) matrix Q = [ q[1,1] q[1,2] .... q[1,n] ] [ . ] [ . ] [ . ] [ q[n-1,1] .... q[n-1,n] ] I can guarantee that all (n-1)x(n-1) matrix minors are non-singular. I'm looking for the general solution of this matrix for a proof I'm working on. Any ideas how to go about doing this? Thanks, Scott -------------------- Isn't this just the vector of the determinants of the minors, times an alternating sign? In[183]:= {{a, b, c, d}, {e, f, g, h}, {i, j, k, l}}; In[184]:= Table[Drop[Transpose[%], {z}], {z, 4}] Out[184]= {{{b, f, j}, {c, g, k}, {d, h, l}}, {{a, e, i}, {c, g, k}, {d, h, l}}, {{a, e, i}, {b, f, j}, {d, h, l}}, {{a, e, i}, {b, f, j}, {c, g, k}}} In[185]:= Det /@ % Out[185]= {-d g j + c h j + d f k - b h k - c f l + b g l, -d g i + c h i + d e k - a h k - c e l + a g l, -d f i + b h i + d e j - a h j - b e l + a f l, -c f i + b g i + c e j - a g j - b e k + a f k} In[186]:= FullSimplify[#*Denominator[#[[1, 1]]] &@NullSpace[%%%]] Out[186]= {{-d g j + c h j + d f k - b h k - c f l + b g l, d g i - c h i - d e k + a h k + c e l - a g l, -d f i + b h i + d e j - a h j - b e l + a f l, c f i - b g i - c e j + a g j + b e k - a f k}} --rwg